An airplane begins its take-off run at A with zero velocity and a constant acceleration a. Knowing that it becomes airborne 30 s later at B with a take-off velocity of 270 km/h, determine (a) the acceleration a, (b) distance AB.

The airplane starts with zero velocity (v0 = 0) and its velocity after 30 s is 270 km/h (converted into m/s: 270 km/h · 1000 m/1 km · 1 h /3600 s = 75 m/s). Then, we can solve the equation to obtain the acceleration:

75 m/s = a · 30 s

75 m/s / 30 s = a

a = 2.5 m/s²

The acceleration of the airplane is 2.5 m/s².

b)The distance AB can be calculated using the equation of position of an object moving in a straight line with constant acceleration:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time "t".

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

If we place the origin of the frame of reference at A, then, x0 = 0. Since the airplane is initially at rest, v0 = 0. So, the equation gets reduced to this:

x = 1/2 · a · t²

Let´s find the position of the airplane after 30 s:

x = 1/2 · 2.5 m/s² · (30 s)²

x = 1125 m

The position of B is 1125 m away from A (the origin), then, the distance AB is 1125 m.

ConcepcionPetillo ConcepcionPetillo · Answer 2 · 2022-03-01T12:42:56+01:00

(a) the acceleration of the plane is 2.5 m/s²

(b) The distance AB is 1125m

Equations of motion:

(a) Given that the initial velocity of the plane at point A is u = 0

and the final velocity of the plane at point B of take-off is v = 270 km/h

[tex]v=\frac{270\times1000}{3600}m/s=75m/s[/tex]

the time taken to teach point B is t = 30s

So from the first equation of motion, we get:

[tex]v=u+at[/tex]

where a is the acceleration.

[tex]75=a\times30\\\\a=2.5\;m/s^2[/tex]

(b) From the second equation of motion we get,

[tex]s=ut+\frac{1}{2} at^2[/tex]

where s is the distance

[tex]s = \frac{1}{2}\times2.5\times30^2\\\\s=1125m[/tex]

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